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8090 [49]
2 years ago
8

Which formula is comparing a sample of a gas under two different conditions of volume and temperature, if the pressure is consta

nt?
⚪︎ V1 V2

____ = ___

T1 T2


⚪︎ xy = k


⚪︎ V

__ = k

T


⚪︎ P1V1 = P2V2


⚪︎ PV = k
Chemistry
1 answer:
Gwar [14]2 years ago
3 0

Answer:

V1/T1 = V2/T2

Explanation:

The Combined gas law for two states, 1 and 2, is given by:

P1V1/T1 = P2V2/T2

where P, V, and T are Pressure, Volume, and Temperature (always in degrees Kelvin).

If pressure is constant, the P1 and P2 cancel out:

V1/T1 = V2/T2

I can't tell from the formatting of the answers if A is this relationship.  Please compare.

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3 years ago
Write the reaction when solid lead(ii) nitrate is put into water:
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Solid lead nitrate in water gives lead oxide and nitric acid

Pb(NO3)2 + H2O ---> PbO + 2 HNO3
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An 11.75 g sample of a common hydrate of cobalt(ii) chloride is heated. after heating, 9.25 g of anhydrous cobalt chloride remai
Irina-Kira [14]
Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt 
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number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol 
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3 0
3 years ago
1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)
Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0
3 years ago
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